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Chapter 9: Problem 75

Consider the test of \(H_{0}: \sigma^{2}=5\) against \(H_{1}: \sigma^{2}<5 .\) Approximate the \(P\) -value for each of the following test statistics. (a) \(x_{0}^{2}=25.2\) and \(n=20\) (b) \(x_{0}^{2}=15.2\) and \(n=12\) (c) \(x_{0}^{2}=4.2\) and \(n=15\)

Short Answer

Expert verified
(a) 0.217; (b) 0.233; (c) < 0.01.

Step by step solution

01

Identify the Test and Distribution

We are conducting a hypothesis test concerning the population variance. Given that the test is for variance, we are using a Chi-square test. Under the null hypothesis \( H_0: \sigma^2 = 5 \), the test statistic \( x_0^2 \) follows a Chi-square distribution with \( n-1 \) degrees of freedom.
02

Calculate Degrees of Freedom

For each case provided, we can calculate the degrees of freedom as \( df = n - 1 \).(a) For \( n = 20 \), \( df = 20 - 1 = 19 \).(b) For \( n = 12 \), \( df = 12 - 1 = 11 \).(c) For \( n = 15 \), \( df = 15 - 1 = 14 \).
03

Determine P-Value from Chi-Square Distribution

Using the Chi-square distribution table or a calculator, determine the \( P \)-value for each test statistic. Given \( H_1: \sigma^2 < 5 \), we need the upper tail (right-tail) probability.(a) For \( x_0^2 = 25.2 \) with \( df = 19 \), find \( P(X^2 \geq 25.2) \).(b) For \( x_0^2 = 15.2 \) with \( df = 11 \), find \( P(X^2 \geq 15.2) \).(c) For \( x_0^2 = 4.2 \) with \( df = 14 \), find \( P(X^2 \geq 4.2) \).
04

Interpret P-Value Results

Using reference materials, compute each \( P \)-value:(a) The \( P \)-value for \( x_0^2 = 25.2 \) with \( df = 19 \) is approximately 0.217.(b) The \( P \)-value for \( x_0^2 = 15.2 \) with \( df = 11 \) is approximately 0.233.(c) The \( P \)-value for \( x_0^2 = 4.2 \) with \( df = 14 \) is less than 0.01 (often reported as \(< 0.01\)).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
Hypothesis testing is a statistical method used to decide if there is enough evidence to support a specific hypothesis about a population parameter. In our exercise, we test the hypothesis about the population variance. The null hypothesis (\( H_0\)) states the population variance, \( \sigma^2\), is equal to 5. This is expressed as \( H_0: \sigma^2 = 5\). The alternative hypothesis (\( H_1\)) posits the variance is less than 5, presented as \( H_1: \sigma^2<5\).

To test these hypotheses, we use a test statistic that follows a specific distribution under the null hypothesis. In our case, we calculate a Chi-square test statistic. If this statistic falls in a certain range based on the Chi-square distribution, it suggests that the alternative hypothesis might be true. The strength of our evidence against the null hypothesis is illustrated by the \( P\)-value.
  • A smaller \( P\)-value indicates stronger evidence against \( H_0\).
  • Generally, a \( P\)-value below a chosen significance level (like 0.05) leads to rejecting \( H_0\).
Understanding hypothesis testing is crucial as it supports making data-driven decisions by objectively analyzing statistical data.
Population Variance
Population variance is a measure of how much the data in a population spread out from the population mean. In simpler terms, it tells us how much the individual data points differ from the average. Variance is an essential concept in statistics because it helps us understand the variability within a dataset.

In our exercise, the goal is to determine if the population variance, \( \sigma^2\), is less than the known value of 5. Calculating the test statistic involves understanding the sample variance and comparing it to the hypothesized population variance. We use a Chi-square test precisely because it is designed to analyze variance data.
  • **Sample Variance**: Estimation of variance based on sample data.
  • **Population Variance**: The actual variance of the entire population.
The Chi-square test statistic compares the sample variance (calculated from data) to the hypothesized population variance to examine if any observed differences are statistically significant.
Degrees of Freedom
Degrees of freedom (\( df\)) is an important concept in inferential statistics, representing the number of values that have the freedom to vary. It is tied to the number of independent observations in a dataset. When calculating statistics like variance and using Chi-square tests, degrees of freedom become vital.

In a Chi-square test for variance, the degrees of freedom depends solely on the sample size. Specifically, it is calculated as the sample size minus one (\( df = n - 1\)). This is because at least one parameter is usually estimated from the data, leaving \( n - 1\) pieces of free information.
  • For example, if a dataset contains 20 observations, and we wish to estimate a single parameter, 19 degrees of freedom are used.
  • \( df = n-1 \) must always be ranging with n, to represent accurate Chi-square distribution scaling.
Proper calculation of degrees of freedom is critical because it influences the shape of the Chi-square distribution curve, which in turn affects the \( P\)-value derived from the Chi-square statistic.

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Most popular questions from this chapter

A manufacturer of interocular lenses is qualifying a new grinding machine and will qualify the machine if there is evidence that the percentage of polished lenses that contain surface defects does not exceed \(2 \%\). A random sample of 250 lenses contains six defective lenses. (a) Formulate and test an appropriate set of hypotheses to determine if the machine can be qualified. Use \(\alpha=0.05\). Find the \(P\) -value. (b) Explain how the question in part (a) could be answered with a confidence interval.

Output from a software package is given below: One-Sample Z: Test of \(m u=35=\) vs not \(=35\) The assumed standard deviation \(=1.8\) $$ \begin{array}{lrrrrrr} \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\ \mathrm{x} & 25 & 35.710 & 1.475 & ? & ? & ? \end{array} $$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) Use the normal table and the above data to construct a \(95 \%\) two-sided \(\mathrm{CI}\) on the mean. (d) What would the P-value be if the alternative hypothesis is \(\mathrm{H}_{1}: \mu>35 ?\)

A biotechnology company produces a therapeutic drug whose concentration has a standard deviation of 4 grams per liter. A new method of producing this drug has been proposed, although some additional cost is involved. Management will authorize a change in production technique only if the standard deviation of the concentration in the new process is less than 4 grams per liter. The researchers chose \(n=10\) and obtained the following data in grams per liter. Perform the necessary analysis to determine whether a change in production technique should be implemented. $$ \begin{array}{ll} \hline 16.628 & 16.630 \\ 16.622 & 16.631 \\ 16.627 & 16.624 \\ 16.623 & 16.622 \\ 16.618 & 16.626 \\ \hline \end{array} $$

The impurity level (in ppm) is routinely measured in an intermediate chemical product. The following data were observed in a recent test: $$ \begin{array}{l} 2.4,2.5,1.7,1.6,1.9,2.6,1.3,1.9,2.0,2.5,2.6,2.3,2.0,1.8, \\ 1.3,1.7,2.0,1.9,2.3,1.9,2.4,1.6 \end{array} $$ Can you claim that the median impurity level is less than \(2.5 \mathrm{ppm} ?\) (a) State and test the appropriate hypothesis using the sign test with \(\alpha=0.05 .\) What is the \(P\) -value for this test? (b) Use the normal approximation for the sign test to test \(H_{0}: \widetilde{\mu}=2.5\) versus \(H_{1}: \tilde{\mu}<2.5 .\) What is the \(P\) -value for this test?

The data from Medicine and Science in Sports and Exercise described in Exercise \(8-48\) considered ice hockey player performance after electrostimulation training. In summary, there were 17 players and the sample standard deviation of performance was 0.09 seconds. (a) Is there strong evidence to conclude that the standard deviation of performance time exceeds the historical value of 0.75 seconds? Use \(\alpha=0.05 .\) Find the \(P\) -value for this test. (b) Discuss how part (a) could be answered by constructing a \(95 \%\) one-sided confidence interval for \(\sigma\)
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